Boundary of the Hyperbolic plane homeomorphic to S1

Question

I am reading an introduction to hyperbolic surfaces which defines a topology on the closure of the hyperbolic plane as follows.

We take the usual open sets of $\mathbb{H}^2$, plus one open set $U_P$ for each open half plane $P$ in $\mathbb{H}^2$.

For each point $x$ in $\mathbb{H}^2$, $x$ belongs to $U_P$ if it lies in $P$.

For each point $x$ in the boundary of $\mathbb{H}^2$ (which we take to be an equivalence class of unit speed geodesic rays), $x$ belongs to $U_P$ if every representative ray of the class eventually lies in $P$.

My question is, why is the boundary of $\mathbb{H}^2$ compact under this topology

Any reference which explains the topology of the hyperbolic plane in greater detail would be greatly appreciated.

Sorry, I don't see why the boundary wouldn't be homeomorphic to just the Real line. The set of half planes is in direct correspondance with [0,∞) is it not? — Pitaya, Apr 02 '18 at 18:29
The boundary of the hyperbolic plane is compact, see Poincare disk model. But the real line is not compact. — Dietrich Burde, Apr 02 '18 at 18:48
I think I phrased my original question wrong. What I would like to understand, is why the boundary is compact under this topology (in the upper half-plane model). — Pitaya, Apr 02 '18 at 18:59

score 2 · Accepted Answer · answered Apr 02 '18 at 19:01

Adding sets $U_P$ specifies basis for topology, but not the whole topology.

Let $x$ and $x'$ be two vertices in $\mathbb{H}^2$. Suppose $r : [0, \infty) \rightarrow \mathbb{H}^2$ is a ray with $r(0) = x$. Consider sequence of geodesics $(r_i)_{i =0} ^\infty$, where $r_i$ is the unique geodesic from $x'$ to $r(i)$ parametrized by the arc length.

Let $$r'(t) = \lim_{i \rightarrow \infty} r_i (t)$$ $r'$ is a geodesic ray, which stays a bounded distance away from $r$. Therefore $r'$ and $r$ represent the same point in the boundary.

Two different rays originating from the same point diverge.

Hence there is a correspondence between rays up to an equivalence class and rays with a specified end-point. Rays with a specified end-point can be parametrized by the direction, so they naturally correspond to points of $S^1$.

The basis arising from $U_P$ is a basis for the usual topology of a circle.

score 2 · Answer 2 · answered Apr 03 '18 at 12:39

Your comments indicate that you are concerned about the point $\infty$ in $\mathbb{R} \cup \infty$.

In the hyperbolic plane, each geodesic determines two half planes, one on each side of the geodesic.

In the upper half plane model of the hyperbolic plane, certain geodesics appear as semicircles meeting the real line at right angles. The two half planes of such a geodesic are the "inside" of the semicircle and the "outside" of the semicircle. If $P$ is the half plane outside that semicircle, then the point $\infty$ must be regarded as a point of the basic open set $U_P$.

Boundary of the Hyperbolic plane homeomorphic to S1

2 Answers2